Even Odd Equations Quiz: Solving Equations Using Even-Odd Properties

Since cos is even, cos(−θ) = cosθ = 0. On [−π, π], cosθ = 0 at θ = −π/2 and θ = π/2. Verification: cos(−(−π/2)) = cos(π/2) = 0 ✓ and cos(−π/2) = cos(π/2) = 0 ✓. Note that at θ = ±π, cos(±π) = −1 ≠ 0, so these are not solutions.

The answer is False. Cosine is an even function, so cos(−θ) = cosθ, not −cosθ. The equation cos(−θ) = −cosθ would require cosθ = 0 at every θ, which is clearly false. Confusing the even property of cosine with the odd property of sine is a common error.

The answer is True. Since sin is odd, sin(−θ) = −sinθ. Setting sin(−θ) = a gives −sinθ = a, which is equivalent to sinθ = −a. The two equations therefore have identical solution sets within the same interval. This relationship is useful because it converts an equation in sin(−θ) into a standard sine equation.

Since sin is odd, sin(−θ) = −sinθ. Setting sin(−θ) = sinθ gives −sinθ = sinθ, so 2sinθ = 0, which means sinθ = 0. On [0, 2π], sinθ = 0 at θ = 0, π, and 2π. Option C is incorrect because sin(−θ) = sinθ is not a universal identity — it holds only when sinθ = 0.

Since sin is odd, sin(−θ) = −sinθ = 0, so sinθ = 0. On [−π, π] restricted to θ ≥ 0, the solutions are θ = 0 and θ = π. Verification: sin(−0) = 0 ✓ and sin(−π) = 0 ✓. Option D incorrectly includes θ = π/2, where sin(π/2) = 1 ≠ 0.

Since sin is odd, sin(−θ) = −sinθ. The equation becomes −sinθ = 1/2, so sinθ = −1/2. On [0, 2π], sinθ = −1/2 at θ = 7π/6 (Quadrant III) and θ = 11π/6 (Quadrant IV). Verification: sin(−7π/6) = −sin(7π/6) = −(−1/2) = 1/2 ✓ and sin(−11π/6) = −sin(11π/6) = −(−1/2) = 1/2 ✓.

Since sin is odd, sin(−θ) = −sinθ. The equation becomes −sinθ = −1, so sinθ = 1. On [0, 2π], sinθ = 1 only at θ = π/2. Verification: sin(−π/2) = −sin(π/2) = −1 ✓. Option A (3π/2) is where sinθ = −1, which is a common confusion.

Since cos is even, cos(−θ) = cosθ = −1/2. On [−π, π], cosθ = −1/2 at θ = 2π/3 (Quadrant II) and θ = −2π/3 (Quadrant III). Verification: cos(−2π/3) = cos(2π/3) = −1/2 ✓ and cos(−(−2π/3)) = cos(2π/3) = −1/2 ✓.

The answer is True. Since cos is even, cos(−θ) = cosθ for all θ. Therefore cos(−θ) = a and cosθ = a are identical equations with identical solution sets on any interval. This property means equations involving cos(−θ) require no transformation before solving.

Since sin is odd, sin(−θ) = −sinθ. The equation becomes −sinθ = −1/2, so sinθ = 1/2. On [−π, π], sinθ = 1/2 at θ = π/6 and θ = 5π/6. Verification: sin(−π/6) = −sin(π/6) = −1/2 ✓ and sin(−5π/6) = −sin(5π/6) = −1/2 ✓. Option A is a common error that arises from confusing the solutions to sinθ = 1/2 with the negatives of those solutions.

Since sin is odd, sin(−θ) = −sinθ. The equation becomes −sinθ = −√2/2, so sinθ = √2/2. On [0, 2π], sinθ = √2/2 at θ = π/4 and θ = 3π/4. Verifying directly: sin(−π/4) = −sin(π/4) = −√2/2 ✓ and sin(−3π/4) = −sin(3π/4) = −√2/2 ✓. Option C is a common error that arises from applying an unnecessary extra transformation.

Since tan is odd, tan(−θ) = −tanθ = 0, so tanθ = 0. On [−π, π], tanθ = sinθ/cosθ = 0 when sinθ = 0. This occurs at θ = −π, 0, and π. At all three values, cosθ ≠ 0 (cos(±π) = −1), so tan is defined. Option A is the most common error — forgetting that θ = ±π also satisfy sinθ = 0 within the interval.

Since sin is odd, sin(−θ) = −sinθ. The equation becomes −sinθ = 0, so sinθ = 0. On [0, 2π], sinθ = 0 at θ = 0, π, and 2π. Option B (θ = π/2) is incorrect because sin(π/2) = 1 ≠ 0. Verification: sin(−0) = 0 ✓, sin(−π) = 0 ✓, sin(−2π) = 0 ✓.

Since cos is even, cos(−θ) = cosθ. The equation becomes cosθ = √3/2. On [0, 2π], cosθ = √3/2 at θ = π/6 (Quadrant I) and θ = 11π/6 (Quadrant IV). Verification: cos(−π/6) = cos(π/6) = √3/2 ✓ and cos(−11π/6) = cos(11π/6) = √3/2 ✓.

Since tan is odd, tan(−θ) = −tanθ. The equation becomes −tanθ = −1, so tanθ = 1. On [0, 2π], tanθ = 1 at θ = π/4 (Quadrant I) and θ = 5π/4 (Quadrant III). Verification: tan(−π/4) = −tan(π/4) = −1 ✓ and tan(−5π/4) = −tan(5π/4) = −1 ✓.

Since sin is odd, sin(−θ) = −sinθ. Setting −sinθ = b and dividing both sides by −1 gives sinθ = −b. This transforms the original equation into a standard sine equation that can be solved using reference angles and quadrant rules. Option A is incorrect because it omits the sign change introduced by the odd property.

Since cos is even, cos(−θ) = cosθ. The equation becomes cosθ = −1. On [0, 2π], cosθ = −1 only at θ = π. Verification: cos(−π) = cos(π) = −1 ✓.

The answer is True. Since tanθ = sinθ/cosθ and sin is odd while cos is even: tan(−θ) = sin(−θ)/cos(−θ) = (−sinθ)/cosθ = −(sinθ/cosθ) = −tanθ. This holds at every θ where tanθ is defined, i.e. where cosθ ≠ 0.

Since tan is odd, tan(−θ) = −tanθ. The equation becomes −tanθ = √3, so tanθ = −√3. On [0, 2π], tanθ = −√3 at θ = 5π/6 (Quadrant II) and θ = 11π/6 (Quadrant IV). Verification: tan(−5π/6) = −tan(5π/6) = −(−√3) = √3 ✓ and tan(−11π/6) = −tan(11π/6) = −(−√3) = √3 ✓.

Since cosine is an even function, cos(−θ) = cosθ holds for every value of θ in its domain, not just specific values. The equation cos(−θ) = cosθ is an identity, so every θ in [0, 2π] satisfies it. Options B, C, and D incorrectly restrict the solution to particular angles.